History and Applications of the Natural Logarithm
Date: 03/02/98 at 11:45:35 From: Paula Fornwalt Subject: history of e We have been using e in continuous growth problems, where we are examining the growth potential of natural disease populations. I'm so surprised at how often this number comes up in other applications, though. Where did e come from? Who first derived it? Why is it so common in the field of biology?
Date: 03/02/98 at 12:32:09 From: Doctor Rob Subject: Re: history of e For its history, see the following URL: Earliest Uses of Symbols for Constants, by Jeff Miller http://jeff560.tripod.com/constants.html John Napier, the inventor of logarithms, is credited with discovering this constant. Leonhard Euler is credited with popularizing the use of the letter e for this number. There are two reasons for its frequent appearance in mathematical contexts. The first is that it is the limit, as n grows without bound, of (1 + 1/n)^n. This particular definition lends itself to problems involving continuous compounding of interest. Think of adding 1/n of the current total to that current total n times. The second is that e is the only constant such that the slope of the graph of y = e^x at 0 is 1; or, in other words, the function e^x is the solution of the differential equation dy/dx = y, with initial conditions y = 1 when x = 0. Similarly, the solution of dy/dx = a*y, with initial conditions y = 1 when x = 0, is e^(a*x). This is the biological connection. You are talking about the instantaneous rate of growth of something being proportional to the amount of the something. That something can be a population of bacteria in a culture, or a population of cockroaches in a garbage dump. Here, "a" represents the fraction of them reproducing at any given time x. If "a" is negative, it could represent the fraction of organisms dying at any given time. This is the model for radioactive decay of radium, uranium, and so on. This is also the approximate model for inflation -- exponential growth. The way to calculate e is by using as many terms as necessary for the accuracy you need in the following infinite series: e = 1 + 1 + 1/2 + 1/(2*3) + 1/(2*3*4) + 1/(2*3*4*5) + 1/(2*3*4*5*6) + ... The terms shown above are already enough to obtain a value for e that is roughly 2.718. -Doctor Rob, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum